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A quadratic function is a polynomial of degree 2, hence its equation is of the form f(x) = ax2 + bx + c, where ‘a’ is a non-zero integer and a, b, and c are real numbers. And how to find the equation of a quadratic function given two points

- While three equations are required to discover the three coefficients, several situations may aid in the development of a particular equation. You are, however, requesting “a” quadratic equation.

With a vertical axis of symmetry, the equation is y = ax2 + bx + c, which may alternatively be written as y = a(x – h)2 + k.

Case 1: If the y values of the two points are the same,

The axis of symmetry will then pass through the middle of the section connecting them. If the provided locations are (1, 7) and (5, 7), the midway is determined by averaging the x coordinates. (1 + 5)/2 = 3. The symmetry axis is x = 3.

From the provided points, two equations can be written.

(1, 7) 7 = a(1 – 3)^2 + k or 7 = 4a + k

(5, 7) 7 = a (5 – 3)

^2 + k or 7 = 4a + k

As predicted, we only have one equation with two unknowns, resulting in no solution. However, to get a solution, select a number for k, say -5, and the equation becomes 7 = 4a – 5

12 = 4a

a = 3

The equation is now y = 3(x – 3)2 -5.

This yields y = 9(x2 -6x +9) -5

After that, y = 9×2 -54x + 76.

Case 2: If the y values of the two points are the same,

Example 1 and 7 (3, 31)

31 = a(3)2 + b(3) + c and 7 = a(1)2 + b(1) + c

(1) 7 = a+b+c, and (2) 31 = 9a+3b+c

Taking equation 1 and subtracting it from equation 2

24 = 8a + 2b

divide by two

12 = 4a + b

b = 12 – 4a

Let a = 2 or any other number of your choice.

b = 12 – 4 (2)

b = 4

Putting into equation 1

7 = (2) + (4) + c

c = -1

The equation that connects the two locations is y = 2×2 + 4x – 1.

- An unlimited number of quadratic functions travel between two locations. Because a quadratic function is specified by three parameters, determining it requires three points.

However, we can identify the family of quadratic functions that pass between two locations.

We get f(x)=ax2+bx+c.

as well as two points on f, (k,l) and (m,n). So

l=ak2+bk+c

n=am2+bm+c

Our unknowns are a, b, and c, and we only have two equations. It’s simplest to get rid of c by subtracting equations:

l−n=a(k2−m2)+b(k−m)

As a result, if km, any quadratic equations passing between the two locations will have parameters a and b that lie on a line. We may make a the independent parameter and calculate b, c, and hence f for each a.

b=l−n−a(k2−m2)

k−m=l−nk−m−a(k+m)

c=l−ak2−bk

I could replace in the value for b, but it doesn’t seem to simplify things very well, so let’s assume you choose a, calculate b, then compute c.

A quadratic function has the formula f(x) = ax2 + bx + c, where a, b, and c are integers and an is not equal to zero. A parabola is a curve that represents the graph of a quadratic function.

The Quadratic Formula is used to solve a quadratic problem.

In standard form, write the quadratic equation as ax2 + bx + c = 0. Determine the values of a, b, and c.

Make a quadratic formula. Then enter the values of a, b, and c.

Simplify.

Examine the solutions.

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